L 2 : x = s + 1, y = s, z = 4s Suppose that C has coordinates (x, y, z). Then from the vector equality AC = BD, one has

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1 The line L through the points A and B is parallel to the vector AB = 3, 2, and has parametric equations x = 3t + 2, y = 2t +, z = t Therefore, the intersection point of the line with the plane should satisfy: 2(3t + 2) 3(2t + ) + 4( t ) = 3 Solving the last equation with respect to the parameter t, we obtain t = 4 and the coordinates of the intersection point are (3( 4) + 2, 2( 4) +, ( 4) ) = ( 0, 7, 3) 2 (a) The curves meet at the point C if and only if components of the vectors r (t) and r 2 (t) coincide, ie t = 3 s, t = s 2, 3 + t 2 = s 2 From the first and last equations, we have 3 + (3 s) 2 = s 2, solving which we obtain s = 2 Now from the first equation t = which also satisfy the second equation The conclusion is that the point (, 0, 4) is the point of intersection of the curves (b) Let s find the tangent vectors of the curves: for C the tangent vector is r (t) =,, 2t and for the curve C 2 the tangent vector is r (s) =,, 2s At the intersection of C and C 2 one has t = and s = 2, so L is parallel to the vector,, 2 and L 2 is parallel to r (s) =,, 4 Since both lines contain the point (, 0, 4), the equations of the lines are and L : x = t +, y = t, z = 2t + 4, L 2 : x = s +, y = s, z = 4s Suppose that C has coordinates (x, y, z) Then from the vector equality AC = BD, one has x 2, y 5, z = 5 3, 2, 3 4, and x = 4, y = 6, z = 6 4 (a) Since AB =,, 2 and AC = 2,,, the scalar projection of AC onto AB is AB AC comp AB AC = = AB ( )( 2) + ( )() + 2() ( ) 2 + ( ) = 3 6 Therefore, the vector projection is this scalar projection times the unit vector in the direction of AB: 3 AB proj AB AC = 6 = AB = /2, /2, AB 2

2 (b) The area S of the triangle ABC is the half of the area of the parallelogram with adjacent sides AB and AC which is the length of the cross product: S = 2 AB AC = , 3, 3 = 2 (c) The line L containing A(2,, 0) and parallel to the vector AB =,, 2 has parametric equations x = t + 2, y = t +, z = 2t The square of the distance from the arbitrary point D on the line to the point C is a function depending on the parameter t: F (t) = CD 2 = ( t + 2) 2 + ( t ) 2 + (2t ) 2 = 6t 2 6t + 6 To minimize the distance, set d dt F (t) = 0, and solve for t to obtain t = /2 Hence, the required distance from the point C to the line L is d = F (/2) = 9/2 5 First, note that any direction vector of the line L is perpendicular both of the normal vectors n = (, 2, ) and n 2 = (2,, ) Hence, L is parallel to the vector v = n n 2 = ( 3,, 5) Now, it is sufficient to find any point on L, ie a point satisfying x 2y+z = and 2x + y + z = Let, for example, x = 0 Then, solving the system of equations 2y + z = and y + z =, we obtain solutions y = 0 and z =, so the point (0, 0, ) is on the line and the parametric equations of L are x = 3t, y = t, z = 5t + 6 The line L is parallel to the vector v =, 4 0, = 2, 2, 0, and L 2 is parallel to the vector v 2 = 4 2, 4 3, 3 ( ) = 2,, 2 We see that the lines L and L 2 are not parallel to each other, because the vectors v and v 2 are not proportional: there is no such a k 0 so that v = k v 2 (one can also check that v v 2 0) The parametric equations of L through (, 0, ) are x = t +, y = 2t, z = and the line L 2 containing (2, 3, ) has parametric equations x = 2s + 2, y = s + 3, z = 2s

3 To find intersection point of L and L 2, we should find values of t and s such that t + = 2s + 2, 2t = s + 3, = 2s Solving the last two equations, we get t = and s = which satisfy the first equation Therefore, the lines L and L 2 intersect at the point (0, 2, ) 7 (a) First, compute the components of the vectors AB, AC, and AD: AB = 3, 4, 2 2 = 2, 3, 4 AC = 4, 3 4, 3 2 = 3,, 5 AD =, 0 4, 2 = 0, 4, 3 The volume V of the parallelepiped is the absolute value of the scalar triple product AB ( AC AB) = = 3 and V is equal to 3 (b) The equation of the plane through A, B, and D is given by the determinant x y 4 z = x y 4 z = 0, or 7x + 6y 8z = (c) The plane through A, B, and C is x y 4 z = x y 4 z = 0, or x 2y +7z 7 = 0 with the normal vector n =, 2, 7 The normal vector to the plane xy is k = (0, 0, ) and cosine of the angle between the planes is cos ϕ = n k n k = (a) The position vector of the particle whose velocity is v(t) and initial position is r 0 = r(t 0 ) can be found as r(t) = r 0 + t t 0 v(t) dt

4 Since t 0 = 0, and r 0 and v are given, one has r(4) =, 5, t, 2 t, dt =, 5, 4 + 6, 32/3, 4 = 5, 47/3, 8 (b) The tangent line to the curve at t = 4 is parallel to the vector v(4) = 8, 4,, and the equation of the line is x = 8t + 5, y = 4t + 47/3, z = t + 8 (c) The position vector of the particle is r(t) = r 0 + t t 0 v(t) dt =, 5, 4 + t 3 /3, 4/3 t 3/2, t, so the particle pass through P : r(9) = 80, 4, 3 (d) The length of the arc traveled from t = to t = 2 is 2 L = (2t) 2 + (2 2 t) dt = (2t + ) dt = 4 9 The surface is a hyperboloid of one sheet x y 2 (/ 3) 2 z 2 (/ 2) 2 = 0 The generic equation of the tangent plane to the graph of z = y ln x at the point where (x, y) = (x 0, y 0 ) is z = y 0 ln x 0 + y 0 x 0 (x x 0 ) + ln x 0 (y y 0 ) and at (, 4, 0) it is z = 4(x 4), or 4x z 4 = 0 To find the distance between the planes, fix any point on the first plane (ie any point A(x, y, z) such that z = 2x + y ) and find a distance from this point to the second plane Let, for example A(, 0, ) Then the distance d from the point A to the plane is d = ()( 4) + 0( 2) + ()(2) 3 ( 4) 2 + ( 2) = 5 24 Another method to solve the problem is the following Find the equations of the line through A perpendicular to the second plane (parallel to it s normal vector): x = 4t +, y = 2t, z = 2t + and the point of intersection with the plane B solving the equation 4( 4t + ) 2( 2t) + 2(2t + ) = 3 for t It s easy to see that t = 5/24 and B(/6, 5/2, 7/2) and d = (/6 ) 2 + ( 5/2 0) 2 + (7/2 ) 2 = 5 24

5 2 The surface is a hyperboloid of one sheet Indeed, by completing the squares, we have and after division by 4: 4 The limit 4x 2 + 4(y 2 2y + ) 4 z 2 = 0, x 2 (y ) z2 2 2 = lim (x,y) (0,0) 3x 2 y 2 2x 4 + y 4 does not exist because if we consider two different directions along two different lines x = 0 and y = x, we obtain different answers: x 2 along x = 0 : y 4 = 0 0 3x 2 x 2 along y = x : 2x 4 + x 4 = 3x4 3x 4 5 It is easy to find two different points on the line by letting t = 0 and t = : Q(2,, 2) and R(5, 0, 4) Now the plane through the points P, Q and R is x y z = 0, or 2x + 4y z = 6 6 (a) f x = 3x 2 y 2, f y = 2xy +, f xy = 2y (b) f x =, f y x 2 +y 2 y =, and f x 2 +y 2 (x+ x 2 +y 2 xy = ) and (c) Let z = f(x, y) be f(x, y) = x 2 cos x 2 y Then f x = 2x cos x 2 y 2x 3 y sin x 2 y, f y = x 4 sin x 2 y, f xy = 4x 3 sin x 2 y 2x 5 y cos x 2 y 7 The linear approximation of f(x, y) at (, ) is y (x 2 +y 2 ) 3/2 L(x, y) = f(, ) + f x (, )(x ) + f y (, )(y ) = e + 2e(x ) + e(y ) since f x = ye x ( + x) and f y = xe x Therefore f(, 09) is approximately L(, 09) = Find the parametrization of the curve: let x = t, then y = t 2 and z = 3t 2, so r(t) = (t, t 2, 2t 2 + t 4 )

1.(6pts) Find symmetric equations of the line L passing through the point (2, 5, 1) and perpendicular to the plane x + 3y z = 9. .(6pts Find symmetric equations of the line L passing through the point (, 5, and perpendicular to the plane x + 3y z = 9. (a x = y + 5 3 = z (b x (c (x = ( 5(y 3 = z + (d x (e (x + 3(y 3 (z = 9 = y 3

Solutions to old Exam 1 problems Solutions to old Exam 1 problems Hi students! I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections

(a) We have x = 3 + 2t, y = 2 t, z = 6 so solving for t we get the symmetric equations. x 3 2. = 2 y, z = 6. t 2 2t + 1 = 0, Name: Solutions to Practice Final. Consider the line r(t) = 3 + t, t, 6. (a) Find symmetric equations for this line. (b) Find the point where the first line r(t) intersects the surface z = x + y. (a) We

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( 1)2 + 2 2 + 2 2 = 9 = 3 We would like to make the length 6. The only vectors in the same direction as v are those 1.(6pts) Which of the following vectors has the same direction as v 1,, but has length 6? (a), 4, 4 (b),, (c) 4,, 4 (d), 4, 4 (e) 0, 6, 0 The length of v is given by ( 1) + + 9 3 We would like to make

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December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in two-dimensional space (1) 2x y = 3 describes a line in two-dimensional space The coefficients of x and y in the equation South Carolina College- and Career-Ready (SCCCR) Pre-Calculus Key Concepts Arithmetic with Polynomials and Rational Expressions PC.AAPR.2 PC.AAPR.3 PC.AAPR.4 PC.AAPR.5 PC.AAPR.6 PC.AAPR.7 Standards Know 1.1.4. Prouct of three vectors. Given three vectors A, B, anc. We list three proucts with formula (A B) C = B(A C) A(B C); A (B C) =B(A C) C(A B); a 1 a 2 a 3 (A B) C = b 1 b 2 b 3 c 1 c 2 c 3 where the